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Assoli18 [71]
3 years ago
14

CAN ANY ONE DO THIS ASAP

Physics
1 answer:
vlabodo [156]3 years ago
8 0

Answer:

v = 75 m / 15 s = 5 m/s

Explanation:

Speed is equal to the distance traveled divided by the time.

v = d / t

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the force between your feet and the floor is greater while standing on your tiptoes than while standing flat on your feet
Elodia [21]
Yes thats correct....becuase all of your weight is concentrated on a small area compared to the larger surface area of your feet!
is that what your question was?
4 0
3 years ago
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The atomic mass of gold is 0.197 kg/mole. how many moles are in 0.566 kg of gold.
iogann1982 [59]

Explanation:

0.566kg *(1mol/0.197 kg)= 2.87 mol gold

note how the units cancel out, if the units do not cancel out (kg/kg=1) then u did something wrong

7 0
3 years ago
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A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught
meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2} (Eq. 1)

Where:

y_{o} - Initial height of the softball, measured in meters.

y - Final height of the softball, measured in meters.

v_{o} - Initial velocity of the softball, measured in meters per second.

t - Time, measured in seconds.

g - Gravitational acceleration, measured in meters per square second.

If we know that y = y_{o}, t = 3.56\,s and g = -9.807\,\frac{m}{s^{2}}, the initial velocity of the softball is:

v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0

3\cdot v_{o} -44.132\,m= 0

v_{o} = 14.711\,\frac{m}{s}

The initial velocity of the softball is 14.711 meters per second.

8 0
3 years ago
What is a distraction that a pedestrian may engage in while crossing the street?
goldenfox [79]

Answer:

a dog walking or their phone rings or heard a neighbor talking to them

6 0
3 years ago
ear the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of
sesenic [268]

Answer:

a) V_{2/1}=0.8m/s

b) The second runner will win

c) d = 10.54m

Explanation:

For part (a):

V_{2/1} = V_{2} - V_{1} = 0.8m/s

For part (b) we will calculate the amount of time that takes both runners to cross the finish line:

t_{1} = \frac{X_{1}}{V_{1}}=\frac{250}{3.45}=72.46s

t_{2} = \frac{X_{2}}{V_{2}}=\frac{250+45}{4.25}=69.41s

Since it takes less time to the second runner to cross the finish line, we can say the she won the race.

For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:

X1 = V1*t2 = 239.46m  Since the finish line was 250m away:

d = 250m - 239.46m = 10.54m

6 0
3 years ago
Read 2 more answers
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